$v(t) = (t^4, 2t^3, t^4)$ What is the speed of $v(t)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\sqrt{32t^6 + 9t}$ (Choice B) B $4t^2\sqrt{2t^2 + 3}$ (Choice C) C $4t\sqrt{t^4 + 6t}$ (Choice D) D $2t^2\sqrt{8t^2 + 9}$
Solution: The speed of a parametric curve is the magnitude of its velocity. If $f(t) = (a(t), b(t), c(t))$, then speed is: $\| f'(t) \| = \sqrt{ a'(t)^2 + b'(t)^2 + c'(t)^2 }$ Our position function here is $v(t)$. $\begin{aligned} v'(t) &= (4t^3, 6t^2, 4t^3) \\ \\ \text{speed} &= ||v'(t)|| \\ \\ &= \sqrt{16t^6 + 36t^4 + 16t^6} \\ \\ &= \sqrt{32t^6 + 36t^4} \\ \\ &= 2t^2\sqrt{8t^2 + 9} \end{aligned}$ Therefore, the speed of $v(t)$ is $2t^2\sqrt{8t^2 + 9}$.